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Levi Civita symbol on Curl of Vector cross product

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Use the LC symbol to calculate the following: $$\nabla \times \frac{\vec{m} \times \hat{r}}{r^2}$$

    Where ##\vec{m}## is just a vector, and ##\hat{r}## is the unit radial vector and ##r## is the length of the radial vector.
    2. Relevant equations
    On the Levi Civita symbol: http://folk.uio.no/patricg/teaching/a112/levi-civita/

    3. The attempt at a solution

    I've gotten to ##\epsilon_{kij} \epsilon_{klm} m_l \frac{r^2 \partial_j r_m - r_m \partial_j r^2}{r^4}##, but I don't know how to take the derivatives..
     
  2. jcsd
  3. Oct 3, 2014 #2

    Fredrik

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    That result doesn't look right to me. How did you get it? Note that if ##\hat r=\frac{\vec r}{r}##, then
    $$\frac{\hat r}{r^2}=\frac{\vec r}{r^3}=\bigg(\frac{r_i}{(r_jr_j)^{\frac 3 2}}\bigg) \hat e_i.$$
     
  4. Oct 3, 2014 #3

    RUber

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    On the link you provided, there is a proof for ## A\times B \times C=B(A \cdot C) - C(A \cdot B) ## using the LC method.
    If ##\vec{m}## is just a vector, does that mean constant, i.e. derivatives equal zero?
    That should leave you with something like ## \vec{m} ( \nabla \cdot \frac{\hat{r}}{r^2}) ##
    You could also look at ## \frac{\hat{r}}{r^2}=\frac{1}{x^2+y^2+z^2}\hat{r}## so taking the partial derivatives should not be too much of a challenge.
     
  5. Oct 4, 2014 #4
    Ruber, yes but here you have a curl along with a function (##r^2##) involved.

    But shouldn't it be much easier to perform the derivation in spherical coordinates? This is what's creating the struggle.

    I just wrote up the definition, ordered the resulting equation and used the product rule to get started with the derivations. Note, my ##r_m##'s are actually the ##m##th coordinate of ##\hat{r}##, of not ##\vec{r}##.

    So basically I started with this: $$\epsilon_{ijk} \partial_j \epsilon_{klm} m_l \frac{r_m}{r^2}$$, and used the product rule after ordering to arrive to the equation in the OP.
     
  6. Oct 4, 2014 #5

    Fredrik

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    But the derivatives are with respect to the mth coordinate of ##\vec r##, so with your definition of ##r_m##, you don't have ##\partial_i x_j=\delta_{ij}##. Instead you have to do something like this (where ##x_m## is the mth component of ##\vec r=r\hat r##):
    $$\partial_i r_j =\frac{\partial}{\partial x_i} r_j =\frac{\partial}{\partial x_i}\frac{x_j}{\sqrt{x_kx_k}}.$$
     
  7. Oct 4, 2014 #6
    But since ##\hat{r}## is just a one-component vector, shouldn't it be extremely straightforward to use spherical-coordinates ##\partial_i## on it? Why do you change to xyz coordinates?
     
  8. Oct 4, 2014 #7

    vela

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    It doesn't seem so straightforward to me. In spherical coordinates, ##\vec{r}## depends on all three coordinates in a relatively complicated way whereas in cartesian coordinates you have simply ##\partial_i x_j = \delta_{ij}##.
     
  9. Oct 4, 2014 #8

    Fredrik

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    I never changed coordinates. I just used the definitions. The cross product is defined by ##x\times y=(x\times y)_i e_i=(\varepsilon_{ijk}x_jy_k) e_i##, where the ##e_i## are the standard basis vectors, not the basis vectors associated with the spherical coordinate system. ##\nabla\times f(x)## is defined similarly, as ##\big(\varepsilon_{ijk}\frac{\partial}{\partial x_j}f_k(x)\big) e_i##.

    In this problem, we have
    $$f(x)=m\times\frac{\hat r}{r^2}=m\times\frac{\vec r}{r^3},$$ and therefore
    $$f_k(x)=\varepsilon_{kln}m_l\frac{x_n}{r^3}=\varepsilon_{kln}m_l\frac{x_n}{(x_px_p)^{\frac 3 2}}.$$ I don't know if there's a way to use both spherical coordinates and the Levi-Civita symbol. (Note that the problem requires you to use the latter). If you know such a way, you can certainly try it.
     
  10. Oct 4, 2014 #9
    OK fine I'll just do it as you said. Thanks.
     
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